Convert the curve to an equation in rectangular coordinates:
x=(t+2); y=-2sqrt(t)
x = t+2 ----> t = x-2
y = -2√t
y^2 = 4t ---> t = y^2/4
then y^2/4 = x-2
y^2 = 4x - 8
On what intervals of t is the curve described by the given parametric equation concave up? Concave down?
x=t^2; y=t^(3) + 3t
I am a bit confused on how to solve this...any help/explanations are welcome!! (& greatly appreciated!)
To convert the given curve into an equation in rectangular coordinates (x and y), we need to eliminate the parameter t by expressing t in terms of x and y.
Given:
x = t + 2
y = -2√t
To eliminate t, we'll start by isolating t in the first equation:
x = t + 2
Rearrange the equation:
t = x - 2
Now, substitute this value of t into the second equation:
y = -2√t
y = -2√(x - 2)
Therefore, the equation of the curve in rectangular coordinates is:
y = -2√(x - 2)
Note that this equation represents the relationship between x and y for all points on the curve.